CODEWITHSUNDEEP
windowsbatch-fileif-statement
Roke
Roke

Reputation: 300

Batch if statements based on opening a program

Is it possible to make an if statement for batch based off a user opening a program?

I tried this:

color c
echo off
cls

:props
if open WINWORD.exe (echo a & pause 10 & exit) 
goto props

Doing so will simply post the error 'WINWORD was unexpected at this time.' and kill the command prompt.

What I am trying to achieve is a batch file that will:

  1. Look if anyone is opening WINWORD.exe
  2. Once somebody has opened the file the command prompt will display 'a'
  3. Exit the command prompt after 10 seconds once 'a' was displayed.

Upvotes: 0

Views: 69

Answers (1)

aschipfl
aschipfl

Reputation: 34919

You can search for the process using tasklist.exe, together with find to count the instances:

set PROC=winword.exe
tasklist /FI "IMAGENAME eq %PROC%" | find /C /I "%PROC%"

I used a variable %PROC% to specify the process (*.exe) name just for the sake of convenience.

find will also exit with the return code (ErrorLevel) of 1 if no instances have been found. With a simple if statement you can do what you requested:

if not ErrorLevel 1 ((echo a) & timeout /T 10 /NOBREAK & exit)

I replaced the pause 10 by the timeout command because pause waits for the user to press any key (any arguments are ignored). The switch /NOBREAK means to ignore any key presses.

The parenthesis around echo a avoids a trailing space to be echoed as well.

So all together, the following should do what you asked for:

color c
echo off
set PROC=winword.exe
cls

:props
tasklist /FI "IMAGENAME eq %PROC%" | find /C /I "%PROC%" > nul
if not ErrorLevel 1 ((echo a) & timeout /T 10 /NOBREAK & exit)
timeout /T 1 /NOBREAK > nul
goto props

The (optional) > nul portion after find prevents it from displaying the found number of instances.

The second delay time timeout /T 1 has been inserted to avoid massive CPU load within this loop structure.

Upvotes: 2

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